Question: The scores on a statewide chemistry exam were normally distributed with $\mu = 71$ and $\sigma = 8$. What fraction of test-takers had a grade between $59$ and $81$ on the exam? Use the cumulative z-table provided below. z.00.01.02.03.04.05.06.07.08.09 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
A cumulative z-table shows the probability that a standard normal variable will be less than a certain value (z) In order to use the z-table, we first need to determine the z-scores of the two grades which we were given. Recall that we can calculate the z-scores by subtracting the mean $(\mu)$ from each exam grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}} = \dfrac{59 - {71}}{{8}} = -1.50} $ $ { z = \dfrac{x - {\mu}}{{\sigma}} = \dfrac{81 - {71}}{{8}} = 1.25} $ We can find the percentage of test-takers that earned between $59$ and $81$ by finding the area between $-1.50$ and $1.25$ under the standard normal curve. After looking up these two z-scores in our z-table, subtracting the two table values will provide us with the total area. Look up $1.25$ on the z-table. This value, $ 0.8944$ , represents the portion of the population that scored lower than $81$ on the exam. Since the normal curve is symmetrical, the area less than $-1.50$ is equal to the area greater than $1.50$ , which can be found by looking up $1.50$ on the z-table and subtracting the table value from $1$ , the total area under the curve. $1 - 0.9332 = 0.0668$ . This value, $0.0668$ , represents the portion of the population that scored lower than $59$ on the exam. Finally, subtract the two cumulative areas to arrive at our final answer: ${ 0.8944 } - { 0.0668 } = 0.8276$ Thus, $82.76\%$ of the test-takers scored between $59$ and $81$ on the chemistry exam.